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Consider Hull White model $dr(t)=[\theta(t)-\alpha(t)r(t)]dt+\sigma(t)dW(t)$ when we solve the SDE above we have $r(t)=e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t})+\sigma e^{-\alpha t}\int_{0}^{t}e^{\alpha u}dW(u) $ and when we take expectation and variance we have $r(t) \sim N(e^{-\alpha t}r(0)+\frac{\theta}{\alpha}(1-e^{-\alpha t}),\frac{\sigma^2}{2\alpha}(1-e^{-\alpha t}))$.

I know the calculate how find SDE and find expectation or variance but I don't understand why $r(t)$ has normal distribution.

thanks.

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For simplicity, We assume that $\alpha$ is a positive constant. You need to show that, for any $t>0$, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed. Consider the process $\{X_t, t \geq 0\}$, where \begin{align*} X_t = \frac{1}{\sqrt{\frac{1}{t}\int_0^t e^{2\alpha u} du}}\int_0^t e^{\alpha u} dW_u, \end{align*} for $t>0$, and $X_0=0$. Then $\{X_t, t \geq 0\}$ is a continuous martingale. Moreover, we have the quadratic variation $\langle X, X\rangle_t = t$. By Levy's characterization, $\{X_t, t > 0\}$ is a Brownian motion. That is, $X_t$ is normally distributed. Consequently, \begin{align*} \int_0^t e^{\alpha u} dW_u \end{align*} is normally distributed, and $r_t$ is also normally distributed.

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This is a special case of the question of why $$ \int_0^T f(t) dW_t $$ is normally distributed for a continuous function $f(t).$ This Ito integral can be approximated by a sum $$ \sum_{i=0}^{N-1} f(i T/N) (W_{(i+1)T/N} - W_{i T/N}) .$$ The Brownian increments $(W_{(i+1)T/N} - W_{i T/N})$ are independent normally distributed random variables. The key point is that the sum of independent normally distributed variables is again normally distributed.

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    $\begingroup$ The second key point is that if a sequence of normal variable converges in law then the limit is also normal. $\endgroup$ – AFK May 18 '15 at 21:17
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    $\begingroup$ The sum, which is normal, converges to the ito integral in $L^2(P)$, and the key point is that the $L^2$ limit of a sequence of normal random variables is still normal. $\endgroup$ – Gordon May 18 '15 at 22:36

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